Data.Set

A set of values a.

$O(1)$. The empty set.

$O(1)$. Create a singleton set.

$O(n\log n)$. Create a set from a list of elements.

If the elements are ordered, a linear-time implementation is used, with the performance equal to fromDistinctAscList.

$O(n)$. Build a set from an ascending list in linear time. The precondition (input list is ascending) is not checked.

$O(n)$. Build a set from a descending list in linear time. The precondition (input list is descending) is not checked.

Since: containers-0.5.8

$O(n)$. Build a set from an ascending list of distinct elements in linear time. The precondition (input list is strictly ascending) is not checked.

$O(n)$. Build a set from a descending list of distinct elements in linear time. The precondition (input list is strictly descending) is not checked.

$O(2^n\log n)$. Calculate the power set of a set: the set of all its subsets.

t `member` powerSet s == t `isSubsetOf` s

Example:

powerSet (fromList [1,2,3]) =
  fromList $ map fromList [[],[1],[1,2],[1,2,3],[1,3],[2],[2,3],[3]]

Since: containers-0.5.11

$O(\log n)$. Insert an element in a set. If the set already contains an element equal to the given value, it is replaced with the new value.

$O(\log n)$. Delete an element from a set.

$O(\log n)$ (alterF f x s) can delete or insert x in s depending on whether an equal element is found in s.

In short:

member x <$> alterF f x s = f (member x s)

Note that unlike insert, alterF will not replace an element equal to the given value.

Note: alterF is a variant of the at combinator from Control.Lens.At.

Since: containers-0.6.3.1

$O(\log n)$. Is the element in the set?

$O(\log n)$. Is the element not in the set?

$O(\log n)$. Find largest element smaller than the given one.

lookupLT 3 (fromList [3, 5]) == Nothing
lookupLT 5 (fromList [3, 5]) == Just 3

$O(\log n)$. Find smallest element greater than the given one.

lookupGT 4 (fromList [3, 5]) == Just 5
lookupGT 5 (fromList [3, 5]) == Nothing

$O(\log n)$. Find largest element smaller or equal to the given one.

lookupLE 2 (fromList [3, 5]) == Nothing
lookupLE 4 (fromList [3, 5]) == Just 3
lookupLE 5 (fromList [3, 5]) == Just 5

$O(\log n)$. Find smallest element greater or equal to the given one.

lookupGE 3 (fromList [3, 5]) == Just 3
lookupGE 4 (fromList [3, 5]) == Just 5
lookupGE 6 (fromList [3, 5]) == Nothing

$O(1)$. Is this the empty set?

$O(1)$. The number of elements in the set.

$O(m\log (\frac{n+1}{m+1})),m\le n$. (s1 `isSubsetOf` s2) indicates whether s1 is a subset of s2.

s1 `isSubsetOf` s2 = all (`member` s2) s1
s1 `isSubsetOf` s2 = null (s1 `difference` s2)
s1 `isSubsetOf` s2 = s1 `union` s2 == s2
s1 `isSubsetOf` s2 = s1 `intersection` s2 == s1

$O(m\log (\frac{n+1}{m+1})),m\le n$. (s1 `isProperSubsetOf` s2) indicates whether s1 is a proper subset of s2.

s1 `isProperSubsetOf` s2 = s1 `isSubsetOf` s2 && s1 /= s2

$O(m\log (\frac{n+1}{m+1})),m\le n$. Check whether two sets are disjoint (i.e., their intersection is empty).

disjoint (fromList [2,4,6])   (fromList [1,3])     == True
disjoint (fromList [2,4,6,8]) (fromList [2,3,5,7]) == False
disjoint (fromList [1,2])     (fromList [1,2,3,4]) == False
disjoint (fromList [])        (fromList [])        == True

xs `disjoint` ys = null (xs `intersection` ys)

Since: containers-0.5.11

$O(m\log (\frac{n+1}{m+1})),m\le n$. The union of two sets, preferring the first set when equal elements are encountered.

The union of the sets in a Foldable structure : (unions == foldl union empty).

$O(m\log (\frac{n+1}{m+1})),m\le n$. Difference of two sets.

Return elements of the first set not existing in the second set.

difference (fromList [5, 3]) (fromList [5, 7]) == singleton 3

$O(m\log (\frac{n+1}{m+1})),m\le n$. See difference.

$O(m\log (\frac{n+1}{m+1})),m\le n$. The intersection of two sets. Elements of the result come from the first set, so for example

import qualified Data.Set as S
data AB = A | B deriving Show
instance Ord AB where compare _ _ = EQ
instance Eq AB where _ == _ = True
main = print (S.singleton A `S.intersection` S.singleton B,
              S.singleton B `S.intersection` S.singleton A)

prints (fromList [A],fromList [B]).

$O(nm)$. Calculate the Cartesian product of two sets.

cartesianProduct xs ys = fromList $ liftA2 (,) (toList xs) (toList ys)

Example:

cartesianProduct (fromList [1,2]) (fromList ['a','b']) =
  fromList [(1,'a'), (1,'b'), (2,'a'), (2,'b')]

Since: containers-0.5.11

$O(n+m)$. Calculate the disjoint union of two sets.

 disjointUnion xs ys = map Left xs `union` map Right ys

Example:

disjointUnion (fromList [1,2]) (fromList ["hi", "bye"]) =
  fromList [Left 1, Left 2, Right "hi", Right "bye"]

Since: containers-0.5.11

$O(n)$. Filter all elements that satisfy the predicate.

$O(\log n)$. Take while a predicate on the elements holds. The user is responsible for ensuring that for all elements j and k in the set, j < k ==> p j >= p k. See note at spanAntitone.

takeWhileAntitone p = fromDistinctAscList . takeWhile p . toList
takeWhileAntitone p = filter p

Since: containers-0.5.8

$O(\log n)$. Drop while a predicate on the elements holds. The user is responsible for ensuring that for all elements j and k in the set, j < k ==> p j >= p k. See note at spanAntitone.

dropWhileAntitone p = fromDistinctAscList . dropWhile p . toList
dropWhileAntitone p = filter (not . p)

Since: containers-0.5.8

$O(\log n)$. Divide a set at the point where a predicate on the elements stops holding. The user is responsible for ensuring that for all elements j and k in the set, j < k ==> p j >= p k.

spanAntitone p xs = (takeWhileAntitone p xs, dropWhileAntitone p xs)
spanAntitone p xs = partition p xs

Note: if p is not actually antitone, then spanAntitone will split the set at some unspecified point where the predicate switches from holding to not holding (where the predicate is seen to hold before the first element and to fail after the last element).

Since: containers-0.5.8

$O(n)$. Partition the set into two sets, one with all elements that satisfy the predicate and one with all elements that don't satisfy the predicate. See also split.

$O(\log n)$. The expression (split x set) is a pair (set1,set2) where set1 comprises the elements of set less than x and set2 comprises the elements of set greater than x.

$O(\log n)$. Performs a split but also returns whether the pivot element was found in the original set.

$O(1)$. Decompose a set into pieces based on the structure of the underlying tree. This function is useful for consuming a set in parallel.

No guarantee is made as to the sizes of the pieces; an internal, but deterministic process determines this. However, it is guaranteed that the pieces returned will be in ascending order (all elements in the first subset less than all elements in the second, and so on).

Examples:

splitRoot (fromList [1..6]) ==
  [fromList [1,2,3],fromList [4],fromList [5,6]]

splitRoot empty == []

Note that the current implementation does not return more than three subsets, but you should not depend on this behaviour because it can change in the future without notice.

Since: containers-0.5.4

$O(\log n)$. Lookup the index of an element, which is its zero-based index in the sorted sequence of elements. The index is a number from 0 up to, but not including, the size of the set.

isJust   (lookupIndex 2 (fromList [5,3])) == False
fromJust (lookupIndex 3 (fromList [5,3])) == 0
fromJust (lookupIndex 5 (fromList [5,3])) == 1
isJust   (lookupIndex 6 (fromList [5,3])) == False

Since: containers-0.5.4

$O(\log n)$. Return the index of an element, which is its zero-based index in the sorted sequence of elements. The index is a number from 0 up to, but not including, the size of the set. Calls error when the element is not a member of the set.

findIndex 2 (fromList [5,3])    Error: element is not in the set
findIndex 3 (fromList [5,3]) == 0
findIndex 5 (fromList [5,3]) == 1
findIndex 6 (fromList [5,3])    Error: element is not in the set

Since: containers-0.5.4

$O(\log n)$. Retrieve an element by its index, i.e. by its zero-based index in the sorted sequence of elements. If the index is out of range (less than zero, greater or equal to size of the set), error is called.

elemAt 0 (fromList [5,3]) == 3
elemAt 1 (fromList [5,3]) == 5
elemAt 2 (fromList [5,3])    Error: index out of range

Since: containers-0.5.4

$O(\log n)$. Delete the element at index, i.e. by its zero-based index in the sorted sequence of elements. If the index is out of range (less than zero, greater or equal to size of the set), error is called.

deleteAt 0    (fromList [5,3]) == singleton 5
deleteAt 1    (fromList [5,3]) == singleton 3
deleteAt 2    (fromList [5,3])    Error: index out of range
deleteAt (-1) (fromList [5,3])    Error: index out of range

Since: containers-0.5.4

$O(\log n)$. Take a given number of elements in order, beginning with the smallest ones.

take n = fromDistinctAscList . take n . toAscList

Since: containers-0.5.8

$O(\log n)$. Drop a given number of elements in order, beginning with the smallest ones.

drop n = fromDistinctAscList . drop n . toAscList

Since: containers-0.5.8

$O(\log n)$. Split a set at a particular index.

splitAt !n !xs = (take n xs, drop n xs)

$O(n\log n)$. map f s is the set obtained by applying f to each element of s.

It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y

$O(n)$. The

mapMonotonic f s == map f s, but works only when f is strictly increasing. The precondition is not checked. Semi-formally, we have:

and [x < y ==> f x < f y | x <- ls, y <- ls]
                    ==> mapMonotonic f s == map f s
    where ls = toList s

$O(n)$. Fold the elements in the set using the given right-associative binary operator, such that foldr f z == foldr f z . toAscList.

For example,

toAscList set = foldr (:) [] set

$O(n)$. Fold the elements in the set using the given left-associative binary operator, such that foldl f z == foldl f z . toAscList.

For example,

toDescList set = foldl (flip (:)) [] set

$O(n)$. A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

$O(n)$. A strict version of foldl. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

$O(n)$. Fold the elements in the set using the given right-associative binary operator. This function is an equivalent of foldr and is present for compatibility only.

Please note that fold will be deprecated in the future and removed.

$O(\log n)$. The minimal element of a set.

Since: containers-0.5.9

$O(\log n)$. The maximal element of a set.

Since: containers-0.5.9

$O(\log n)$. The minimal element of a set.

$O(\log n)$. The maximal element of a set.

$O(\log n)$. Delete the minimal element. Returns an empty set if the set is empty.

$O(\log n)$. Delete the maximal element. Returns an empty set if the set is empty.

$O(\log n)$. Delete and find the minimal element.

deleteFindMin set = (findMin set, deleteMin set)

$O(\log n)$. Delete and find the maximal element.

deleteFindMax set = (findMax set, deleteMax set)

$O(\log n)$. Retrieves the maximal key of the set, and the set stripped of that element, or Nothing if passed an empty set.

$O(\log n)$. Retrieves the minimal key of the set, and the set stripped of that element, or Nothing if passed an empty set.

$O(n)$. An alias of toAscList. The elements of a set in ascending order. Subject to list fusion.

$O(n)$. Convert the set to a list of elements. Subject to list fusion.

$O(n)$. Convert the set to an ascending list of elements. Subject to list fusion.

$O(n)$. Convert the set to a descending list of elements. Subject to list fusion.

$O(n\log n)$. Show the tree that implements the set. The tree is shown in a compressed, hanging format.

$O(n\log n)$. The expression (showTreeWith hang wide map) shows the tree that implements the set. If hang is True, a hanging tree is shown otherwise a rotated tree is shown. If wide is True, an extra wide version is shown.

Set> putStrLn $ showTreeWith True False $ fromDistinctAscList [1..5]
4
+--2
|  +--1
|  +--3
+--5

Set> putStrLn $ showTreeWith True True $ fromDistinctAscList [1..5]
4
|
+--2
|  |
|  +--1
|  |
|  +--3
|
+--5

Set> putStrLn $ showTreeWith False True $ fromDistinctAscList [1..5]
+--5
|
4
|
|  +--3
|  |
+--2
   |
   +--1

$O(n)$. Test if the internal set structure is valid.